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The question is: A service organization in a large town organizes a raffle. One thousand raffle tickets are sold for $\$1$ each. Each has an equal chance of winning. First prize is $\$300$, second prize is $\$200$, and third prize is $\$100$. What is that chance of winning something?

First way: $X= \{100, 200, 300\}$, $P(X) = P(100)+ P(200)+ P(300)$ as there is one such ticket for each => $P(X) = 0.001 + 0.001 + 0.001 = 0.003$

Second. when the results are going to be announced, the chance for each ticket to be the first prize winner is $0.001$. to be the second prize winner the chance is, $0.999$ (not to be the first prize winner + $1/999$ (for being the second) in the same way the chance of winning the 3rd position is $998/999 \times1/998$.

=> $P(X) = 0.001 + 999/1000 \cdot 1/999 + 998/1000 + 1/998 =0.003001\dots$

Does watching the show where they announce the names of the winners increase the chance of winning? :D

Kenta S
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nima
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3 Answers3

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The second way of calculating is more complex than needed. If you practicize it correctly then you get the same outcome:

$P\left(W_{1}\right)=\frac{1}{1000}$

$P\left(W_{2}\right)=P\left(W_{1}^{\complement}\cap W_{2}\right)=P\left(W_{1}^{\complement}\right)P\left(W_{2}\mid W_{1}^{\complement}\right)=\frac{999}{1000}\frac{1}{999}=\frac{1}{1000}$

$P\left(W_{3}\right)=P\left(W_{1}^{\complement}\cap W_{2}^{\complement}\cap W_{3}\right)=P\left(W_{1}^{\complement}\cap W_{2}^{\complement}\right)P\left(W_{3}\mid W_{1}^{\complement}\cap W_{2}^{\complement}\right)=\frac{998}{1000}\frac{1}{998}=\frac{1}{1000}$

$P\left(W_{1}\cup W_{2}\cup W_{3}\right)=P\left(W_{1}\right)+P\left(W_{2}\right)+P\left(W_{3}\right)=\frac{3}{1000}$

Here $W_i$ denotes the event of winning the $i$-th prize.

drhab
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  • thank you very much for your thorough answer. I had forgotten to multiple the W∁1 for the third place:( 999/1000 * 998/999 * 1/998) – nima Jul 20 '20 at 17:31
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Let $F$ be the event that you win first prize, and $S$ the event that you win second prize. Assuming that first prize is drawn first and second prize second, it's true that $$\Pr(S|F^c)=\frac1{999},$$ but the probability of winning second prize is $$\Pr(F)\Pr(S|F)+\Pr(F^c)\Pr(S|F^c)=0+\frac{999}{1000}\frac1{999}=\frac1{1000}$$

So no, watching doesn't somehow change the probabilities.

saulspatz
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If watching the show changed the probabilities how would you explain that watching in real time rather than with a tape delay increased the value of the ticket you bought in advance?

Related answer: Is this lot drawing fair?

Ethan Bolker
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